3.96.64 \(\int \frac {-10+2 e^3-2 x+e^x (-50+e^3 (10-10 x)+35 x+10 x^2)+e^x (10-7 x-2 x^2+e^3 (-2+2 x)) \log (x)+(-50+10 e^3-15 x+(10-2 e^3+3 x) \log (x)) \log (5-\log (x))}{e^x (25 x-5 e^3 x+5 x^2)+e^x (-5 x+e^3 x-x^2) \log (x)+(25 x-5 e^3 x+5 x^2+(-5 x+e^3 x-x^2) \log (x)) \log (5-\log (x))} \, dx\)

Optimal. Leaf size=28 \[ \log \left (\frac {\left (e^x+\log (5-\log (x))\right )^2}{\left (-5+e^3-x\right ) x^2}\right ) \]

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Rubi [F]  time = 5.25, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-10+2 e^3-2 x+e^x \left (-50+e^3 (10-10 x)+35 x+10 x^2\right )+e^x \left (10-7 x-2 x^2+e^3 (-2+2 x)\right ) \log (x)+\left (-50+10 e^3-15 x+\left (10-2 e^3+3 x\right ) \log (x)\right ) \log (5-\log (x))}{e^x \left (25 x-5 e^3 x+5 x^2\right )+e^x \left (-5 x+e^3 x-x^2\right ) \log (x)+\left (25 x-5 e^3 x+5 x^2+\left (-5 x+e^3 x-x^2\right ) \log (x)\right ) \log (5-\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10 + 2*E^3 - 2*x + E^x*(-50 + E^3*(10 - 10*x) + 35*x + 10*x^2) + E^x*(10 - 7*x - 2*x^2 + E^3*(-2 + 2*x))
*Log[x] + (-50 + 10*E^3 - 15*x + (10 - 2*E^3 + 3*x)*Log[x])*Log[5 - Log[x]])/(E^x*(25*x - 5*E^3*x + 5*x^2) + E
^x*(-5*x + E^3*x - x^2)*Log[x] + (25*x - 5*E^3*x + 5*x^2 + (-5*x + E^3*x - x^2)*Log[x])*Log[5 - Log[x]]),x]

[Out]

2*x - 2*Log[x] - Log[5 - E^3 + x] + 2*Defer[Int][1/(x*(-5 + Log[x])*(E^x + Log[5 - Log[x]])), x] + 10*Defer[In
t][Log[5 - Log[x]]/((-5 + Log[x])*(E^x + Log[5 - Log[x]])), x] - 2*Defer[Int][(Log[x]*Log[5 - Log[x]])/((-5 +
Log[x])*(E^x + Log[5 - Log[x]])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10 \left (1-\frac {e^3}{5}\right )-2 x+e^x \left (-50+e^3 (10-10 x)+35 x+10 x^2\right )+e^x \left (10-7 x-2 x^2+e^3 (-2+2 x)\right ) \log (x)+\left (-50+10 e^3-15 x+\left (10-2 e^3+3 x\right ) \log (x)\right ) \log (5-\log (x))}{x \left (5-e^3+x\right ) (5-\log (x)) \left (e^x+\log (5-\log (x))\right )} \, dx\\ &=\int \left (\frac {-2 \left (5-e^3\right )+\left (7-2 e^3\right ) x+2 x^2}{x \left (5-e^3+x\right )}-\frac {2 (-1-5 x \log (5-\log (x))+x \log (x) \log (5-\log (x)))}{x (-5+\log (x)) \left (e^x+\log (5-\log (x))\right )}\right ) \, dx\\ &=-\left (2 \int \frac {-1-5 x \log (5-\log (x))+x \log (x) \log (5-\log (x))}{x (-5+\log (x)) \left (e^x+\log (5-\log (x))\right )} \, dx\right )+\int \frac {-2 \left (5-e^3\right )+\left (7-2 e^3\right ) x+2 x^2}{x \left (5-e^3+x\right )} \, dx\\ &=-\left (2 \int \left (-\frac {1}{x (-5+\log (x)) \left (e^x+\log (5-\log (x))\right )}-\frac {5 \log (5-\log (x))}{(-5+\log (x)) \left (e^x+\log (5-\log (x))\right )}+\frac {\log (x) \log (5-\log (x))}{(-5+\log (x)) \left (e^x+\log (5-\log (x))\right )}\right ) \, dx\right )+\int \left (2+\frac {1}{-5+e^3-x}-\frac {2}{x}\right ) \, dx\\ &=2 x-2 \log (x)-\log \left (5-e^3+x\right )+2 \int \frac {1}{x (-5+\log (x)) \left (e^x+\log (5-\log (x))\right )} \, dx-2 \int \frac {\log (x) \log (5-\log (x))}{(-5+\log (x)) \left (e^x+\log (5-\log (x))\right )} \, dx+10 \int \frac {\log (5-\log (x))}{(-5+\log (x)) \left (e^x+\log (5-\log (x))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 30, normalized size = 1.07 \begin {gather*} -2 \log (x)-\log \left (5-e^3+x\right )+2 \log \left (e^x+\log (5-\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + 2*E^3 - 2*x + E^x*(-50 + E^3*(10 - 10*x) + 35*x + 10*x^2) + E^x*(10 - 7*x - 2*x^2 + E^3*(-2 +
2*x))*Log[x] + (-50 + 10*E^3 - 15*x + (10 - 2*E^3 + 3*x)*Log[x])*Log[5 - Log[x]])/(E^x*(25*x - 5*E^3*x + 5*x^
2) + E^x*(-5*x + E^3*x - x^2)*Log[x] + (25*x - 5*E^3*x + 5*x^2 + (-5*x + E^3*x - x^2)*Log[x])*Log[5 - Log[x]])
,x]

[Out]

-2*Log[x] - Log[5 - E^3 + x] + 2*Log[E^x + Log[5 - Log[x]]]

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fricas [A]  time = 0.53, size = 28, normalized size = 1.00 \begin {gather*} -\log \left (x - e^{3} + 5\right ) - 2 \, \log \relax (x) + 2 \, \log \left (e^{x} + \log \left (-\log \relax (x) + 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*exp(3)+3*x+10)*log(x)+10*exp(3)-15*x-50)*log(5-log(x))+((2*x-2)*exp(3)-2*x^2-7*x+10)*exp(x)*lo
g(x)+((-10*x+10)*exp(3)+10*x^2+35*x-50)*exp(x)+2*exp(3)-2*x-10)/(((x*exp(3)-x^2-5*x)*log(x)-5*x*exp(3)+5*x^2+2
5*x)*log(5-log(x))+(x*exp(3)-x^2-5*x)*exp(x)*log(x)+(-5*x*exp(3)+5*x^2+25*x)*exp(x)),x, algorithm="fricas")

[Out]

-log(x - e^3 + 5) - 2*log(x) + 2*log(e^x + log(-log(x) + 5))

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giac [A]  time = 0.25, size = 28, normalized size = 1.00 \begin {gather*} -\log \left (x - e^{3} + 5\right ) - 2 \, \log \relax (x) + 2 \, \log \left (e^{x} + \log \left (-\log \relax (x) + 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*exp(3)+3*x+10)*log(x)+10*exp(3)-15*x-50)*log(5-log(x))+((2*x-2)*exp(3)-2*x^2-7*x+10)*exp(x)*lo
g(x)+((-10*x+10)*exp(3)+10*x^2+35*x-50)*exp(x)+2*exp(3)-2*x-10)/(((x*exp(3)-x^2-5*x)*log(x)-5*x*exp(3)+5*x^2+2
5*x)*log(5-log(x))+(x*exp(3)-x^2-5*x)*exp(x)*log(x)+(-5*x*exp(3)+5*x^2+25*x)*exp(x)),x, algorithm="giac")

[Out]

-log(x - e^3 + 5) - 2*log(x) + 2*log(e^x + log(-log(x) + 5))

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maple [A]  time = 0.06, size = 29, normalized size = 1.04

method	result	size
risch	\(-2 \ln \relax (x )-\ln \left (-{\mathrm e}^{3}+x +5\right )+2 \ln \left ({\mathrm e}^{x}+\ln \left (5-\ln \relax (x )\right )\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*exp(3)+3*x+10)*ln(x)+10*exp(3)-15*x-50)*ln(5-ln(x))+((2*x-2)*exp(3)-2*x^2-7*x+10)*exp(x)*ln(x)+((-10
*x+10)*exp(3)+10*x^2+35*x-50)*exp(x)+2*exp(3)-2*x-10)/(((x*exp(3)-x^2-5*x)*ln(x)-5*x*exp(3)+5*x^2+25*x)*ln(5-l
n(x))+(x*exp(3)-x^2-5*x)*exp(x)*ln(x)+(-5*x*exp(3)+5*x^2+25*x)*exp(x)),x,method=_RETURNVERBOSE)

[Out]

-2*ln(x)-ln(-exp(3)+x+5)+2*ln(exp(x)+ln(5-ln(x)))

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maxima [A]  time = 0.71, size = 28, normalized size = 1.00 \begin {gather*} -\log \left (x - e^{3} + 5\right ) - 2 \, \log \relax (x) + 2 \, \log \left (e^{x} + \log \left (-\log \relax (x) + 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*exp(3)+3*x+10)*log(x)+10*exp(3)-15*x-50)*log(5-log(x))+((2*x-2)*exp(3)-2*x^2-7*x+10)*exp(x)*lo
g(x)+((-10*x+10)*exp(3)+10*x^2+35*x-50)*exp(x)+2*exp(3)-2*x-10)/(((x*exp(3)-x^2-5*x)*log(x)-5*x*exp(3)+5*x^2+2
5*x)*log(5-log(x))+(x*exp(3)-x^2-5*x)*exp(x)*log(x)+(-5*x*exp(3)+5*x^2+25*x)*exp(x)),x, algorithm="maxima")

[Out]

-log(x - e^3 + 5) - 2*log(x) + 2*log(e^x + log(-log(x) + 5))

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mupad [B]  time = 9.95, size = 28, normalized size = 1.00 \begin {gather*} 2\,\ln \left (\ln \left (5-\ln \relax (x)\right )+{\mathrm {e}}^x\right )-\ln \left (x-{\mathrm {e}}^3+5\right )-2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - 2*exp(3) - exp(x)*(35*x + 10*x^2 - exp(3)*(10*x - 10) - 50) + log(5 - log(x))*(15*x - 10*exp(3) -
log(x)*(3*x - 2*exp(3) + 10) + 50) + exp(x)*log(x)*(7*x + 2*x^2 - exp(3)*(2*x - 2) - 10) + 10)/(log(5 - log(x)
)*(25*x - 5*x*exp(3) - log(x)*(5*x - x*exp(3) + x^2) + 5*x^2) + exp(x)*(25*x - 5*x*exp(3) + 5*x^2) - exp(x)*lo
g(x)*(5*x - x*exp(3) + x^2)),x)

[Out]

2*log(log(5 - log(x)) + exp(x)) - log(x - exp(3) + 5) - 2*log(x)

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sympy [A]  time = 0.93, size = 26, normalized size = 0.93 \begin {gather*} - 2 \log {\relax (x )} + 2 \log {\left (e^{x} + \log {\left (5 - \log {\relax (x )} \right )} \right )} - \log {\left (x - e^{3} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*exp(3)+3*x+10)*ln(x)+10*exp(3)-15*x-50)*ln(5-ln(x))+((2*x-2)*exp(3)-2*x**2-7*x+10)*exp(x)*ln(x
)+((-10*x+10)*exp(3)+10*x**2+35*x-50)*exp(x)+2*exp(3)-2*x-10)/(((x*exp(3)-x**2-5*x)*ln(x)-5*x*exp(3)+5*x**2+25
*x)*ln(5-ln(x))+(x*exp(3)-x**2-5*x)*exp(x)*ln(x)+(-5*x*exp(3)+5*x**2+25*x)*exp(x)),x)

[Out]

-2*log(x) + 2*log(exp(x) + log(5 - log(x))) - log(x - exp(3) + 5)

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